---
layout: post
title: An interesting equality from linear algebra
date: '2015-02-06T15:00:00.001-08:00'
author: Alex
tags:
- Math
modified_time: '2015-02-08T08:27:07.663-08:00'
blogger_id: tag:blogger.com,1999:blog-307916792578626510.post-2001257522060513388
blogger_orig_url: http://brilliantlywrong.blogspot.com/2015/02/just-interesting-equality-from-linear.html
---

<p>
    It's a bit mind-blowing when you want to understand this equality geometrically:
    <br/>we have vectors $x_1, ..., x_n$ in n-dimensional space
    <br/>Let's take an  orthonormal basis in this space $e_1, ..., e_n$ and compute such vectors obtained with via scalar product:
</p>
<p>
    $$ z_i = (&lt; e_i, x_1 &gt;, &lt; e_i, x_2 &gt;, ..., &lt; e_i, x_n &gt;) $$
</p>
<p>
    So the theorem is:
    $$ \det_{i,j} &lt; z_i, z_j &gt; = \det_{i,j} &lt; x_i, x_j &gt; $$
</p>
<p>
    And the proof is very simple, let's introduce the matrix $A$: $ A_{ij} = &lt;x_i, e_j&gt; $
    $$ \det_{i,j} &lt; z_i, z_j &gt; = \det A^T A = \det A A^T = \det_{i,j} &lt; x_i, x_j &gt;$$
</p>
<p>
    Voila! Beautiful, but completely unclear.
</p>
